Question

If $x = -5$ is a root of $ \begin{vmatrix}2x+1&4&8\\ 2&2x&2\\ 7&6&2x\end{vmatrix}=0$ then the other roots are

• A. 3, 3.5
• B. 1, 3.5
• C. 1, 7
• D. 2, 7

Answer 1

Answer: (B)

Given : $\begin{vmatrix}2x+1&4&8\\ 2&2x&2\\ 7&6&2x\end{vmatrix}=0$
$\therefore \left(2x+1\right)\left(4x^{2}-12\right)-4\left(4x-14\right)+8\left(12-14x\right)=0$
$\Rightarrow 8x^{3}-24x+4x^{2}-12-16x+56+96-112x = 0$
$\Rightarrow 8x^{3}+4x^{2}-152x+140=0$
$\left(x+5\right)$ is a factor of above equation
$\therefore 8x^{3}+40x^{2}-36x^{2}-180x+28x+140=0$
$\Rightarrow 8x^{2}\left(x+5\right)-36x\left(x+5\right)+28\left(x+5\right)=0$
$\Rightarrow \left(x+5\right)\left(8x^{2}-36x+28\right)=0$
$\Rightarrow \left(x+5\right)4\left(2x^{2}-9x+7\right)=0$
$\Rightarrow 4\left(x+5\right)\left(2x^{2}-7x-2x+7\right)=0$
$\Rightarrow 4\left(x+5\right)\left[x\left(2x-7\right)-1\left(2x-7\right)\right]=0$
$\Rightarrow 4\left(x+5\right)\left(2x-7\right)\left(x-1\right)=0$
$\Rightarrow x=-5,3.5,1$